Question: Simplify and expand the following expression: $ \dfrac{5r - 2}{5r + 6}-\dfrac{r - 2}{5r + 7} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5r + 6)(5r + 7)$ Multiply the first term by $\dfrac{5r + 7}{5r + 7}$ $ \begin{align*} \dfrac{5r - 2}{5r + 6} \times \dfrac{5r + 7}{5r + 7} & = \dfrac{(5r - 2)(5r + 7)}{(5r + 6)(5r + 7)} \\ & = \dfrac{25r^2 + 25r - 14}{(5r + 6)(5r + 7)}\end{align*} $ Multiply the second term by $\dfrac{5r + 6}{5r + 6}$ $ \begin{align*} \dfrac{r - 2}{5r + 7} \times \dfrac{5r + 6}{5r + 6} & = \dfrac{(r - 2)(5r + 6)}{(5r + 7)(5r + 6)} \\ & = \dfrac{5r^2 - 4r - 12}{(5r + 7)(5r + 6)}\end{align*} $ Now we have: $ = \dfrac{25r^2 + 25r - 14}{(5r + 6)(5r + 7)} - \dfrac{5r^2 - 4r - 12}{(5r + 7)(5r + 6)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{25r^2 + 25r - 14 - (5r^2 - 4r - 12)}{(5r + 6)(5r + 7)} $ $ = \dfrac{25r^2 + 25r - 14 - 5r^2 + 4r + 12}{(5r + 6)(5r + 7)} $ $ = \dfrac{20r^2 + 29r - 2}{(5r + 6)(5r + 7)}$ Expand the denominator: $ = \dfrac{20r^2 + 29r - 2}{25r^2 + 65r + 42}$